I tried using this method.
Let a be x, b be y+3
(a+b)^2 + (a-b)^2 = 0
2a^2 + 2b^2 = 0
(x)^2 + (y+3)^2 = 0
But if I bring x^2 over,
(y+3)^2 = -x^2
I doubt there is such curve and if there is, it is undefined isn't it?

You are right, it isn't much of a graph. The left side of your last equation is non-negative and the right side is non-positive. The only way they can be equal is if both sides are zero. Can you find any (x,y) that does that? If so, whatever you find constitutes your graph.